Examples of PDE

Standard

1. PDE Exercises

f: \mathbb{R}^n \Rightarrow \mathbb{C} given {\tilde{M}f} is a solution of the homogeneous Cauchy problem, with {\tilde{M}f(0,x) = 0}, {\frac{\partial \tilde{M}f}{\partial t}(0,x) = f(x)} Let {g:[0,\infty) \times \mathbb{R}^n \Rightarrow \mathbb{C}} be sufficiently often continuously differentiable. Show without using any explicit formular for {\tilde{M}f} that the function Qg defined by Qg(t,x):= \int^{t}_{0} \tilde{M}g_s(t-s,x)ds, where {g_s(x):= g(s,x),} is the solution of the Cauchy problem for the inhomogeneous wave equation with right hand side g and zero initial conditions. We want to show that this implies {\Box Qg = g} {Qg(0,x)= \frac{\partial Qg}{\partial t}(0,x) = 0} \newline For n = 1 this gives an alternative solution of Example 1. \tilde{M}f(0,x) = \frac{1}{2c} \int^{x+t}_{x-t}f(\tau)d\tau This implies

\displaystyle  Qg(t,x) = \frac{1}{2c} \int^{t}_{0} \int^{x+c(t-s)}_{x-c(t-s)}g(s,\tau)d\tau ds \ \ \ \ \ (1)

We can then write Qg(0,x) = {\int_0^0 \cdots ds = 0} {\frac{\partial Q}{\partial t}(t,x) = \tilde{M} g_s(t-s,x)} We let s = t {+\int^t_0 \frac{\partial}{\partial t} \tilde{M}g_s(t-s,x)ds} We know {\tilde{M}g_t(0,x) =0} SO:

\displaystyle  \frac{\partial Qg}{\partial t}(t,x) = \int^{t}_0 \frac{\partial}{\partial t} \tilde{M}g_s(t-s,x)ds \ \ \ \ \ (2)

In particular if t=0 We get {\frac{\partial Qg}{\partial t} (0,x) = \int^0_0 \cdots = 0} We also have to look at the box operator {\Box = \frac{\partial^2}{\partial t^2} - \triangle_x} we obtain by applying \frac{\partial}{\partial t}.

By parameter dependent integrals; \frac{\partial^2 Qg}{\partial t^2}(t,x) = \int^{t}_0 \frac{\partial^2}{\partial t^2} \tilde{M}g_s(t-s,x) + \frac{\partial}{\partial t} \tilde{M}g_s (t-s,x)|_{s=t}
\frac{\partial^2 Qg(t,x)}{\partial t^2} = g(t,x) + \int^t_0 \cdots ds By parameter dependent integrals; \triangle_x Qg(t,x) =\int^t_0 \triangle_x \tilde{M}g_s(t-s,x) ds \Box Qg(t,x) = g(t,x) + \int^t_0 \Box Mg(t-s,x)ds The integral above goes to zero by the homogeneous Cauchy problem solution. Therefore we can write

\displaystyle  \Box Qg(t,x) = g(t,x)



We only need the properties of {\tilde{M}g_s}. This should be seen as an exercise in parameter dependent integrals.

2. Examples of distributions and their order

Let us recall the criterion: T:{C^{\infty}_{c}(U) \Rightarrow \mathbb{C}} linear T is a distribution {\iff} {\forall K \subset U} compact {\exists p(K) \in \mathbb{N}_0}, {C(K) \in [0,\infty)} such that for {\phi \in C^{\infty}_{c}(U)} with supp{\phi \subset K} |T(\phi)| \leq C(K) \sum_{|\alpha|\leq p(K)}\|D^{\alpha}\phi\|_{\infty} Where the last norm is the suprenum norm. \delta_{x_{0}} |\delta_{x_{0}}| = |\phi(x_{0})| \leq \|\phi \|_{\infty} (c=1, p = 0, indepedent of K. f \in L^1_{loc}(U) {T_f(\phi) = \int_U f(x) \phi (x) ds} {|T_f(\phi)| = |\int_U f(x) \phi(x) dx| \leq \int_U|f(x)||\phi(x)|dx}

Assume {K \subset U} compact {supp \phi \subset K} {=\int_K|f(x)||\phi(x)|dx} {\leq \|\phi\|_\infty \int_K |f(x)|dx} the integral above can be considered as C(K) and it is less than infinity (p=0, independent of K). {T(\phi) = \int_M \phi|_M . \omega} M {\subset U} k-dim oriented subset and we can consider {\omega} a k(top) form There is a trick:

In local coordinates {Y_1,\cdots Y_k} {V\subset M} open {y_1: V \Rightarrow \mathbb{R}} {\omega = g(y_1,\cdots, y_k)dY_1\wedge \cdots \wedge dY_k} We use the pullback in fact, but we can sloppy write \int_V \phi|_M \omega = \int_V \phi(Y_1 \cdots Y_k)g(Y_1,\cdots,Y_k)dy_1,\cdots, dy_k {\int_M \phi_M \omega =} sum over open sets, where we have coordinates using partition of unity. We need a volume form or take the modulus of {|g(y_1,...,y_k)|} If we define, in local coordinates of the correct orientation {|\omega(x)|=|g(y_1,\cdots,y_k)|dy_1 \wedge \cdots \wedge dy_k} then {|\omega|} is well-defined independent of co-ordinates. We can then write the following |\int_M \phi|_M . \omega| \leq \int_M|\phi|_M . |\omega| = \int_{K \cap M} |\phi|_{|M}.|\omega| \leq \|\phi\|_\infty . \int_{K \cap M} |\omega| Because {supp \phi \subset K} we can do this trick. The integral above becomes C(K) and is finite, since {\omega} has a finite number. This is indeed a distribution and clearly of order 0.

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